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Let $R$ be a commutative ring and $A_1$, $A_2$ be $R$-algebras. Is there any general mean to compute the Jacobson radical of the tensor product $A_1\otimes_R A_2$ in terms of ${\rm Rad}(A_i )$, $i=1,2$?

I am mainly interested in the case where $R$ is the ring of integers of a $p$-adic field $F$ and the $A_i$'s are hereditary orders in some simple $F$-algebras.

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    $\begingroup$ If $L/K$ is a purely inseparable extension of fields, the radical of $L\otimes _KL$ is nonzero while that of $L$ is of course zero. This leaves little hope to find a general expression for the radical of a tensor product. $\endgroup$
    – abx
    Commented Nov 7, 2018 at 20:45
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    $\begingroup$ In general I think there is nothing useful known, but in some special situations we have $rad(A_1 \otimes_R A_2)=rad(A_1) \otimes_R A_2 + A_1 \otimes_R rad(A_2)$. For example when $A_i$ are finite dimensional over a seperable field R. $\endgroup$
    – Mare
    Commented Nov 7, 2018 at 21:01
  • $\begingroup$ @Mare what do you call a separable field? a perfect field? $\endgroup$
    – YCor
    Commented Nov 7, 2018 at 23:01
  • $\begingroup$ @YCor Yes, I meant perfect. $\endgroup$
    – Mare
    Commented Nov 8, 2018 at 7:41

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